PowerPoint PresentationSlide 2Damped LCR OscillatorSlide 4Slide 5SummarySlide 7Slide 8Slide 9Slide 10Slide 11Slide 12Slide 13Slide 14Slide 15Slide 16Slide 17Slide 18Physics 2102Jonathan DowlingLecture 21: THU 01 APR 2010Lecture 21: THU 01 APR 2010Ch. 31.4–7: Electrical Oscillations, Ch. 31.4–7: Electrical Oscillations, LC Circuits, Alternating CurrentLC Circuits, Alternating CurrentQuickTime™ and a decompressorare needed to see this picture.QuickTime™ and a decompressorare needed to see this picture.QuickTime™ and a decompressorare needed to see this picture.LC Circuit: At t=0 1/3 Of Energy Utotal is on Capacitor C and Two Thirds On Inductor L. Find Everything! (Phase 0=?) UB(0)UE0( )=12L q0ωsin(ϕ0)⎡⎣⎤⎦212Cq0cos(ϕ0)⎡⎣⎤⎦2=Utotal/ 32Utotal/ 3 UB(0)UE0( )=12L q0ωsin(ϕ0)⎡⎣⎤⎦212Cq0cos(ϕ0)⎡⎣⎤⎦2=Utotal/ 32Utotal/ 3UB(t) =12L q0ωsin(ωt +ϕ0)[ ]2UEt( )=12Cq0cos(ωt +ϕ0)[ ]2UB(t) =12L q0ωsin(ωt +ϕ0)[ ]2UEt( )=12Cq0cos(ωt +ϕ0)[ ]2UB(0) =12L q0ωsin(ϕ0)[ ]2=Utotal/ 3UE0( )=12Cq0cos(ϕ0)[ ]2=2Utotal/ 3UB(0) =12L q0ωsin(ϕ0)[ ]2=Utotal/ 3UE0( )=12Cq0cos(ϕ0)[ ]2=2Utotal/ 3 LC q0ωsin(ϕ0)⎡⎣⎤⎦2q0cos(ϕ0)⎡⎣⎤⎦2=12 LC q0ωsin(ϕ0)⎡⎣⎤⎦2q0cos(ϕ0)⎡⎣⎤⎦2=12 ω =1/ LCq0=VC ω =1/ LCq0=VC tan(ϕ0) =12ϕ0=arctan 1/ 2( )=35.3° tan(ϕ0) =12ϕ0=arctan 1/ 2( )=35.3°q =q0cos(ωt +ϕ0)i(t) =−q0ωsin(ωt +ϕ0)′i (t) =−ω2q0cos(ωt +ϕ0)VL(t) =−q0Ccos(ωt +ϕ0)VC(t) =q0Ccos(ωt +ϕ0)q =q0cos(ωt +ϕ0)i(t) =−q0ωsin(ωt +ϕ0)′i (t) =−ω2q0cos(ωt +ϕ0)VL(t) =−q0Ccos(ωt +ϕ0)VC(t) =q0Ccos(ωt +ϕ0)Damped LCR OscillatorDamped LCR OscillatorIdeal LC circuit without resistance: oscillations go on forever; = (LC)–1/2Real circuit has resistance, dissipates energy: oscillations die out, or are “damped”Math is complicated! Important points:–Frequency of oscillator shifts away from = (LC)-1/2 –Peak CHARGE decays with time constant = QLCR=2L/R–For small damping, peak ENERGY decays with time constant ULCR= L/RCRL0 4 8 12 16 200.00.20.40.60.81.0Etime (s)Umax=Q22Ce−RtLU22If we add a resistor in an circuit (see figure) we mustmodify the energy equation, because now energy isbeing dissipated on the resistor: .E BRLdUi RdtqU U U=−= + =Damped Oscillations in an CircuitRCL222 2Li dU q dq diLi i RC dt C dt dt+ → = + =−( )222/222 210. This is the same equation as thatof the damped harmonics osc 0, which has theillator: The angular f solution re( ) c que:os .bt mmd x dxm b kxdt dtxdq di d q d q dqi L R qdt dt dt dt dtt x e tCω φ−= → = → + + =+ + =′= +( )2/2222ncy For the damped circuit the solution is:The angular freque1 ( ) cos . .4ncy . 4Rt LRqk bRCt Qe tmLLC Lmω φ ωω−′ ′−−+== =′(31-6)/ 2Rt LQe−/ 2Rt LQe−( )q tQQ−( )q t( )/ 2( ) cosRt Lq t Qe tω φ−′= +2214RLC Lω′= −/222The equations above describe a harmonic oscillator with an exponentially decayingamplitude . The angular frequency of the damped oscillator1 is always smaller than the angular 4Rt LQeRLC Lω−′= −221frequency of the 1undamped oscillator. If the term we can use the approximation .4LCRL LCωω ω=′≈=τRC= RCτRL= L / RτRCL= 2L / RSummarySummary•Capacitor and inductor combination produces an electrical oscillator, natural frequency of oscillator is =1/√LC•Total energy in circuit is conserved: switches between capacitor (electric field) and inductor (magnetic field).•If a resistor is included in the circuit, the total energy decays (is dissipated by R).Alternating Current:To keep oscillations going we need to drive the circuit with an external emf that produces a current that goes back and forth.Notice that there are two frequencies involved: one at which the circuit wouldoscillate “naturally”. The other is the frequency at which we drive theoscillation. However, the “natural” oscillation usually dies off quickly (exponentially) with time. Therefore in the long run, circuits actually oscillate with the frequency at which they are driven. (All this is true for the gentleman trying to make the lady swing back and forth in the picture too).We have studied that a loop of wire,spinning in a constant magnetic fieldwill have an induced emf that oscillates with time, € E =Emsin(ωdt)That is, it is an AC generator.Alternating Current:AC’s are very easy to generate, they are also easy to amplify anddecrease in voltage. This in turn makes them easy to send in distributiongrids like the ones that power our homes. Because the interplay of AC and oscillating circuits can be quitecomplex, we will start by steps, studying how currents and voltagesrespond in various simple circuits to AC’s.AC Driven Circuits:1) A Resistor:0=−Rvemf € vR= emf = Emsin(ωdt) € iR=vRR=EmRsin(ωdt)Resistors behave in AC very much as in DC, current and voltage are proportional (as functions of time in the case of AC),that is, they are “in phase”.For time dependent periodic situations it is useful torepresent magnitudes using Steinmetz “phasors”. These are vectors that rotate at a frequency d , their magnitude is equal to the amplitude of the quantity in question and their projection on the vertical axis represents the instantaneous value of the quantity.QuickTime™ and a decompressorare needed to see this picture.CharlesSteinmetzAC Driven Circuits:2) Capacitors: € vC= emf = Emsin(ωdt) € qC= C emf = CEmsin(ωdt) € iC=dqCdt= ωdCEmcos(ωdt) € iC= ωdCEmsin(ωdt + 900) € iC=EmXsin(ωdt + 900)reactance"" 1 whereCXdω= € im=EmX€ looks like i =VRCapacitors “oppose a resistance” to AC (reactance) of frequency-dependent magnitude 1/d C (this idea is true only for maximum amplitudes, the instantaneous story is more complex).AC Driven Circuits:3) Inductors: € vL= emf = Emsin(ωdt)LdtvidtidLvLLLL∫=⇒= € iL= −EmLωdcos(ωdt) € =EmLωdsin(ωdt − 900) € iL=EmXsin(ωdt − 900) € im=EmXdLX ω= whereInductors “oppose a resistance” to AC (reactance) of frequency-dependent magnitude d L(this idea is true only for maximum amplitudes, the instantaneous story is more complex).Power Station Transmission lines Erms =735 kV , I rms = 500 A Home110 VT1T2Step-up transformerStep-down transformerR = 220Ω1000 km=l2heat rmsThe resistance of the power line . is fixed (220 in our example). Heating
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